Models with Memory

Jesse Brunner

A simple example

Imagine we measured the relative size of six individual…plants? frogs?

Have measurements at weeks 0, 2, 3, 4, 6, and 8

Some data are missing

Simulating our example

n <- 6
time <- c(0, 2, 4, 6, 8)
betas <- rnorm(n=n, mean = 0.9, sd = 0.25)
sigma <- 0.75

df <- expand.grid(ID = 1:n,
                  time = time)
df$size <- rnorm(n=nrow(df), 
                 mean=betas[df$ID]*df$time, 
                 sd=sigma)
df$size[df$time == 0] <- 0
df <- df[-sample(1:nrow(df), 10), ]

Simulating our example

Analysis goal

Our research question is whether plants or frogs grow substantially over time.

(And probably we’d have different treatments, but we’re keeping it simple so far.)

How would you analyze these data?

Option 1: ignore the grouping… complete pooling

individuals are essentially identical

m1 <- ulam(
  alist(
    size ~ dnorm(mu, sigma), 
    mu <- beta*time,
    
    # priors
    beta ~ dnorm(1,1),
    sigma ~ dexp(2)
  ), data = df
)
precis(m1)
           mean         sd      5.5%     94.5%    n_eff    Rhat4
beta  0.7817148 0.05464689 0.7027511 0.8676466 304.5726 0.998089
sigma 1.1559751 0.18563845 0.8952203 1.4708731 332.8428 1.003157

Option 1: ignore the grouping… complete pooling

Option 2: every individual (or group) is different… no pooling

m2 <- ulam(
  alist(
    size ~ dnorm(mu, sigma), 
    mu <- beta[ID]*time,
    
    # priors
    beta[ID] ~ dnorm(1, 1),
    sigma ~ dexp(2)
  ), data = df
)
precis(m2, depth=2)
             mean         sd      5.5%     94.5%    n_eff     Rhat4
beta[1] 0.6255864 0.07320952 0.5112366 0.7375515 612.3183 1.0011711
beta[2] 1.0211894 0.06275966 0.9159211 1.1224503 505.3025 0.9985341
beta[3] 0.5695921 0.28890830 0.1000682 1.0489905 879.6874 1.0001581
beta[4] 0.9185527 0.05630575 0.8372090 1.0010603 561.9530 0.9982647
beta[5] 0.4354022 0.06661095 0.3252638 0.5414363 618.9719 1.0013894
beta[6] 0.8040047 0.09282635 0.6572829 0.9610849 441.2600 0.9979987
sigma   0.6329067 0.12479079 0.4662342 0.8700223 380.4414 0.9982784

Option 2: every individual (or group) is different… no pooling

Option 2: every individual (or group) is different… no pooling

Option 2b: average across group estimates

mean across individuals estimates

   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
 0.5712  0.6916  0.7288  0.7291  0.7593  0.8912

standard deviation among individual estimates

   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
 0.1551  0.2192  0.2453  0.2552  0.2811  0.4988

Option 3: partial pooling

every individual (or group) is different, but not that different

m3 <- ulam(
  alist(
    size ~ dnorm(mu, sigma), 
    mu <- beta[ID]*time,
    
    # priors
    beta[ID] ~ dnorm(beta_mu, beta_sd),
    beta_mu ~ dnorm(1,1),
    beta_sd ~ dexp(1),
    sigma ~ dexp(2)
  ), data = df
)
precis(m3)
             mean        sd      5.5%     94.5%    n_eff     Rhat4
beta_mu 0.7350791 0.1424153 0.4985496 0.9628633 342.0578 0.9997341
beta_sd 0.2984620 0.1399021 0.1432614 0.5721289 364.6213 1.0083107
sigma   0.6394945 0.1406827 0.4642217 0.8840324 399.8765 1.0020758

Option 3: partial pooling

Option 3: partial pooling

slightly changed predictions

Predictions are drawn towards the mean, all else equal.